Passing variables in URL

ExeOutput for PHP
1

Is it possible to pass variables to my opening php file?, for example…
/index.php?var1=fred&var2=smith
Thank you.

2

Sure, place your full URL here:

3

Hi. I’m trying to do this exactly. I have the following in the “Display the following URL…” setting and I do not get anything when I try to extract the variable. I’m using
ghe://heserver/index.php?token=hello and in my php index.php file I have the following:
var_dump($argv) and it displays NULL, changing to http://heserver/index.php?token=hello also gives me NULL back.

Was anyone able to get this working?

4

$argv is for PHP command-line programs. You must use the _REQUEST variable.

5

So I was wrong to use var_dump($argv), I was supposed to be using var_dump($_GET), it is working.

But can I get access to information like token=%COMPUTERNAME% ?

6

What is %COMPUTERNAME%? It’s not defined by ExeOutput. But you could also take a look at PHP’s variables (just run phpinfo()).